What is the difference between using $1 vs \1 in Perl regex substitutions?

Question

I\'m debugging some code and wondered if there is any practical difference between $1 and \\1 in Perl regex substitutions

For example:

my $package_n         


        

Answer 1

First, you should always use warnings when developing:

#!/usr/bin/perl

use strict; use warnings;

my $package_name = "Some::Package::ButNotThis";

$package_name =~ s{^(\w+::\w+)}{\1};

print $package_name, "\n";

Output:

\1 better written as $1 at C:\Temp\x.pl line 7.

When you get a warning you do not understand, add diagnostics:

C:\Temp> perl -Mdiagnostics x.pl
\1 better written as $1 at x.pl line 7 (#1)
    (W syntax) Outside of patterns, backreferences live on as variables.
    The use of backslashes is grandfathered on the right-hand side of a
    substitution, but stylistically it's better to use the variable form
    because other Perl programmers will expect it, and it works better if
    there are more than 9 backreferences.

Why does it work better when there are more than 9 backreferences? Here is an example:

#!/usr/bin/perl

use strict; use warnings;

my $t = (my $s = '0123456789');
my $r = join '', map { "($_)" } split //, $s;

$s =~ s/^$r\z/\10/;
$t =~ s/^$r\z/$10/;

print "[$s]\n";
print "[$t]\n";

Output:

C:\Temp> x
]
[9]

If that does not clarify it, take a look at:

C:\Temp> x | xxd
0000000: 5b08 5d0d 0a5b 395d 0d0a                 [.]..[9]..

See also perlop:

The following escape sequences are available in constructs that interpolate and in transliterations …

\10 octal is 8 decimal. So, the replacement part contained the character code for BACKSPACE.

NB

Incidentally, your code does not do what you want: That is, it will not print Some::Package some package contrary to what your comment says because all you are doing is replacing Some::Package with Some::Package without touching ::ButNotThis.

You can either do:

($package_name) = $package_name =~ m{^(\w+::\w+)};

or

$package_name =~ s{^(\w+::\w+)(?:::\w+)*\z}{$1};